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I know that if X and Y are random scalar variables, then: \begin{align*} \mathrm{Var}(X+Y) & = \mathrm{Var}(X) + \mathrm{Var}(Y) + 2\mathrm{Corr}(X,Y)\sqrt{\mathrm{Var}(X)\mathrm{Var}(Y)} \\ & \le \left(\sqrt{\mathrm{Var}(X)} + \sqrt{\mathrm{Var}(Y)}\right)^2 \end{align*} I would like to know if the same inequality holds for random vectors, i.e.: $X , Y \in \mathcal{R^{n}}$

\begin{align*} \mathrm{Var}(X+Y) & = \mathrm{Var}(X) + \mathrm{Var}(Y) + \mathrm{Cov}(X,Y)+ \mathrm{Cov}(X,Y)' \\ & \le \left(\sqrt{\mathrm{Var}(X)} + \sqrt{\mathrm{Var}(Y)}\right)\left(\sqrt{\mathrm{Var}(X)} + \sqrt{\mathrm{Var}(Y)}\right)' \end{align*} Where square root of a variance (positive definite) matrix is its Cholesky decomposition. And the inequality means that the difference is semidefinite positive.

As an extra hypothesis, I have that both $\mathrm{Var}(X)$ and $\mathrm{Var}(Y)$ are diagonal matrices.

I have been looking for Cauchy Schwartz inequalities for variance matrices, and I found this: \begin{equation} \mathrm{Var}(Y) \ge \mathrm{Cov}(Y,X) \mathrm{Var}(X)^{-1} \mathrm{Cov}(X,Y) \end{equation} But I don't know how to give any use to it. Thanks a lot!

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This doesn't hold. Take this example: Let $X, \, Y \in \mathcal{R}^2$ and the variance matrix of $\left( \begin{matrix} X \\ Y \end{matrix} \right)$ is $D =\left( \begin{matrix} 250 & 150 & 155 & 115 \\ 150 & 100 & 95 & 80 \\ 155 & 95 & 133 & 87 \\ 115 & 80 & 87 & 70 \\ \end{matrix} \right) $.

Which is positive definite with eigenvalues : 0.0029, 14.3177 29.2519 and 509.4275.

Now, $\mathcal{Var}(X+Y) = \left(\begin{matrix} 693 & 447 \\ 447 &330\end{matrix}\right)$ and

$\sqrt{\mathrm{Var}(X)}=\text{Chol}\left[ \left(\begin{matrix} 250 & 150 \\ 150 & 100 \end{matrix}\right)\right] = \left(\begin{matrix} 15.8114 & 0 \\ 9.4868 & 3.1623 \end{matrix}\right)$ $\sqrt{\mathrm{Var}(Y)}=\text{Chol}\left[ \left(\begin{matrix} 133 & 87 \\ 150 & 100 \end{matrix}\right)\right] = \left(\begin{matrix} 11.5326 & 0 \\ 7.5439 & 3.6180 \end{matrix}\right)$

Sadly, $\left(\sqrt{\mathrm{Var}(X)} + \sqrt{\mathrm{Var}(Y)}\right)\left(\sqrt{\mathrm{Var}(X)} + \sqrt{\mathrm{Var}(Y)}\right)' - \mathrm{Var}(X+Y) = \left(\begin{matrix} 54.6917 & 18.6863 \\ 18.6863 & 6.0171 \end{matrix}\right) $

has -0.3292 as eigenvalue, so it's not positive definite.

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